You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction

An insulated ...

Question

An insulated container of two diatomic gases has two chambers separated by an insulting partition. One of the chambers has volume $V_{1}$ and contains ideal gas at pressure $p_{1}$ and temperature $T_{1}$ . The other chamber has volume $V_{2}$ and contains ideal gas at pressure $p_{2}$ and temperature $T_{2}$ . If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

$\frac{T_{1} T_{2} (p_{1} V_{1} + p_{2} V_{2})}{p_{1} V_{1} T_{1} + p_{2} V_{2} T_{2}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}{p_{1} V_{1} + p_{2} V_{2}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{p_{1} V_{1} T_{1} + p_{2} V_{2} T_{2}}{p_{1} V_{1} + p_{2} V_{2}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{T_{1} T_{2} (p_{1} V_{1} + p_{2} V_{2})}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$\frac{T_{1} T_{2} (p_{1} V_{1} + p_{2} V_{2})}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}$

As no work is done and system is thermally insulated from surrounding, it means sum of internal energy of gas in two partitions is constant i.e., $U = U_{1} + U_{2}$ . Assuming both gases have same degree of freedom, then
$U = \frac{f (n_{1} + n_{2}) R T}{2} a n d U_{1} = \frac{f (n_{1} R T_{1})}{2}, U_{2} = \frac{f n_{2} R T_{2}}{2}$

Solving we get, $T = \frac{(p_{1} V_{1} + p_{2} V_{2}) T_{1} T_{2}}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}$

Suggest Corrections

Similar questions

Q. An insulated container of two diatomic gases has two chambers separated by an insulting partition. One of the chambers has volume

V_{1}

and contains ideal gas at pressure

p_{1}

and temperature

T_{1}

. The other chamber has volume

V_{2}

and contains ideal gas at pressure

p_{2}

and temperature

T_{2}

. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Q. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume

V_{1}

and contains ideal gas at pressure

P_{1}

and temperature

T_{1}

. The other chamber has volume

V_{2}

and contains ideal gas at pressure

P_{2}

and temperature

T_{2}

. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be :

Q. An ideal gas in a thermally insulated container, at internal pressure

P_{1},

volume

V_{1}

and absolute temperature

T_{1}

expands irreversibly against zero external pressure so that the pressure of gas becomes

P_{2},

volume

V_{2}

and temperature

T_{2}

. Which of the following statement(s) is / are incorrect?

Q. The rectangular box shown in Fig has a partition which can slide without friction along the length of the box. Initially, each of the two chambers of the box has one mole of a mono-atomic ideal gas

(γ = 5 / 3)

at a pressure

P_{0}

, volume

V_{0}

and temperature

T_{0}

. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition, until the final pressure in both chambers becomes

243 P_{0} / 32

. Determine the work done by the gas in the right chamber.
[Take

R = 8.3 J/K mol

]

Q. Two chambers, one containing

m_{1} g

of a gas at

P_{1}

pressure and other containing

m_{2} g

of a gas at

P_{2}

pressure are put in communication with each other. If temperature remains constant, the common pressure reached will be:

Join BYJU'S Learning Program